$ C = \left[\begin{array}{rrr}2 & 3 & -1 \\ 0 & -2 & 4\end{array}\right]$ $ D = \left[\begin{array}{rr}2 & 4 \\ 2 & -1 \\ -2 & 5\end{array}\right]$ What is $ C D$ ?
Because $ C$ has dimensions $(2\times3)$ and $ D$ has dimensions $(3\times2)$ , the answer matrix will have dimensions $(2\times2)$ $ C D = \left[\begin{array}{rrr}{2} & {3} & {-1} \\ {0} & {-2} & {4}\end{array}\right] \left[\begin{array}{rr}{2} & \color{#DF0030}{4} \\ {2} & \color{#DF0030}{-1} \\ {-2} & \color{#DF0030}{5}\end{array}\right] = \left[\begin{array}{rr}? & ? \\ ? & ?\end{array}\right] $ To find the element at any row $i$ , column $j$ of the answer matrix, multiply the elements in row $i$ of the first matrix, $ C$ , with the corresponding elements in column $j$ of the second matrix, $ D$ , and add the products together. So, to find the element at row 1, column 1 of the answer matrix, multiply the first element in ${\text{row }1}$ of $ C$ with the first element in ${\text{column }1}$ of $ D$ , then multiply the second element in ${\text{row }1}$ of $ C$ with the second element in ${\text{column }1}$ of $ D$ , and so on. Add the products together. $ \left[\begin{array}{rr}{2}\cdot{2}+{3}\cdot{2}+{-1}\cdot{-2} & ? \\ ? & ?\end{array}\right] $ Likewise, to find the element at row 2, column 1 of the answer matrix, multiply the elements in ${\text{row }2}$ of $ C$ with the corresponding elements in ${\text{column }1}$ of $ D$ and add the products together. $ \left[\begin{array}{rr}{2}\cdot{2}+{3}\cdot{2}+{-1}\cdot{-2} & ? \\ {0}\cdot{2}+{-2}\cdot{2}+{4}\cdot{-2} & ?\end{array}\right] $ Likewise, to find the element at row 1, column 2 of the answer matrix, multiply the elements in ${\text{row }1}$ of $ C$ with the corresponding elements in $\color{#DF0030}{\text{column }2}$ of $ D$ and add the products together. $ \left[\begin{array}{rr}{2}\cdot{2}+{3}\cdot{2}+{-1}\cdot{-2} & {2}\cdot\color{#DF0030}{4}+{3}\cdot\color{#DF0030}{-1}+{-1}\cdot\color{#DF0030}{5} \\ {0}\cdot{2}+{-2}\cdot{2}+{4}\cdot{-2} & ?\end{array}\right] $ Fill out the rest: $ \left[\begin{array}{rr}{2}\cdot{2}+{3}\cdot{2}+{-1}\cdot{-2} & {2}\cdot\color{#DF0030}{4}+{3}\cdot\color{#DF0030}{-1}+{-1}\cdot\color{#DF0030}{5} \\ {0}\cdot{2}+{-2}\cdot{2}+{4}\cdot{-2} & {0}\cdot\color{#DF0030}{4}+{-2}\cdot\color{#DF0030}{-1}+{4}\cdot\color{#DF0030}{5}\end{array}\right] $ After simplifying, we end up with: $ \left[\begin{array}{rr}12 & 0 \\ -12 & 22\end{array}\right] $